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Homework Help: VERY simple Velocity vs time graph interpretation

  1. Jan 25, 2008 #1
    Wow, I'm extremely frustrated with this one.

    1. The problem statement, all variables and given/known data

    http://img262.imageshack.us/img262/4616/probre1.jpg [Broken]

    I forgot to label the axis's. The X is time and the Y is Velocity

    3. The attempt at a solution

    Iv done a few things on this one:

    First, I figured I would just count the area under it but I submitted the answer and it was wrong

    Second, I used x = x(o) + v(o)t plugging in the velocity and time from the graph along with the x(o) which is given in the problem

    Any help is appriciated, Thank You
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 25, 2008 #2
    You have to find the area under the curve to find position.

    If you can't maybe this will help: http://img230.imageshack.us/img230/513/17914212vr1.jpg [Broken]
     
    Last edited by a moderator: May 3, 2017
  4. Jan 25, 2008 #3
    See, I did that and counted the area under the positive area in the first quadrant and then I subtracted it from the negative area in the 4th quad. I got 29 for the positive area and 6 for the negative, giving me 23 and it was wrong. So I figured maybe I had to add the areas for 35 but that as well came out wrong. :confused:
     
  5. Jan 25, 2008 #4
    23m is the displacement. You are asked for the position.
     
  6. Jan 25, 2008 #5
    ughh, 1 more hint, im not getting it
     
  7. Jan 25, 2008 #6
    If the initial position is [itex]x(0)=-10 m[/itex] and the displacement is 15 m, what's the final position?
     
  8. Jan 25, 2008 #7
    x(0) is 10. I'm confused how the position is different from the displacement. I know the displacement is how far away it is from the orgin no matter how it travels but the position would be. I'm just not seeing it .
     
  9. Jan 25, 2008 #8
    You are initial at [itex]x(0)=10 m[/itex]. If your displacement [itex]\Delta x[/itex] is -10 m you should arrive to the origin.

    [tex]\Delta x=x_f-x_i\Rightarrow x_f=\Delta x+x_i[/tex]

    Do you understand that?
     
  10. Jan 25, 2008 #9
    I'm sorry, This isn't making sense to me, I'm usually good with graph interpretation.
     
  11. Jan 25, 2008 #10
    Ok, let's try this.

    The initial position is 12 m. You walk 4 m at the positive direction. What is your displacement and what is your final position?
     
  12. Jan 25, 2008 #11
    displacement would be 4 m and your final position is 16 m. I see where your going with this so I would simply add my displacement to the initial position? make the answer 33?
     
  13. Jan 25, 2008 #12
  14. Jan 25, 2008 #13
    You know what it was? I was thinking of the X axis as if it were distance even though I KNEW it was time. Because of this I was thinking its position was at 9 and from the initial position it would be -1 m . DUHHH. Thanks so much for your help.
     
  15. Jan 25, 2008 #14
    Always it helps if you draw a picture o the actuall movement!
    Glad I helped! :smile:
     
  16. Jan 26, 2008 #15
    When you take the area under a curve you're finding the change in what ever quantity you're asked for. You need some sort of condition if you want to find the exact value. Without knowing that x(0) = 10m the best you could say about the graph would be the object moved 23m but I can't tell you EXACTLY where it is or EXACTLY where is started. Does that make sense?
     
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